Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 20

int x/(16x^4-1)dx
To solve using partial fraction method, the denominator of the integrand should be factored.
x/(16x^4-1)=x/((2x-1)(2x+1)(4x^2+1))
Take note that if the factors in the denominator are linear, each factor has a partial fraction in the form A/(ax+b) .
If the factors are in quadratic form, each factor has a partial fraction in the form (Ax+B)/(ax^2+bx+c) .
So expressing the integrand as sum of fractions, it becomes:
x/((2x-1)(2x+1)(4x^2+1))=A/(2x-1)+B/(2x+1)+(Cx+D)/(4x^2+1)
To determine the values of A, B, C and D, multiply both sides by the LCD of the fractions present.
(2x-1)(2x+1)(4x^2+1)*x/((2x-1)(2x+1)(4x^2+1))=(A/(2x-1)+B/(2x+1)+(Cx+D)/(4x^2+1)) *(2x-1)(2x+1)(4x^2+1)
x= A(2x+1)(4x^2+1) + B(2x-1)(4x^2+1)+ (Cx+D)(2x-1)(2x+1)
Then, assign values to x in which either 2x-1, 2x+1, or4x^2+1 will become zero.
So plug-in x=1/2 to get the value of A.
1/2=A(2(1/2)+1)(4(1/2)^2+1)+B(2(1/2)-1)(4(1/2)^2 + 1) + (C(1/2)+D)(2(1/2)-1)(2(1/2)+1)
1/2=A(4) + B(0)+(C(1/2)+D)(0)
1/2=4A
1/8=A
Plug-in x=-1/2 to get the value of B.
-1/2=A(2(-1/2)+1)(4(-1/2)^2+1) + B(2(-1/2)-1)(4(-1/2)^2+1) + (C(-1/2)+D)(2(-1/2)-1)(2(-1/2)+1)
-1/2=A(0)+B(-4) +(C(-1/2)+D)(0)
-1/2=-4B
1/8=B
To solve for D, plug-in the values of A and B. Also, plug-in x=0.
0=1/8(2(0)+1)(4(0)^2+1) + 1/8(2(0)-1)(4(0)^2+1) + (C(0)+D)(2(0)-1)(2(0)+1)
0=1/8 - 1/8 -D
0=D
To solve for C, plug-in the values of A, B and D. Also, assign any value to x. Let it be x=1.
1=1/8(2(1)+1)(4(1)^2+1) +1/8(2(1) -1)(4(1)^2+1) + (C(1) + 0)(2(1) -1)(2(1)+1)
1=15/8+5/8+C(3)
1=10/4+3C
-3/2=3C
-1/2=C
So the partial fraction decomposition of the integrand is:
int x/(16x^4-1)dx
=int(x/((2x-1)(2x+1)(4x^2+1)))dx
=int (1/(8(2x-1)) + 1/(8(2x+1)) - x/(2(4x^2+1)) )dx
Then, express it as three integrals.
=int 1/(8(2x-1))dx + int 1/(8(2x+1))dx - int x/(2(4x^2+1)) dx
=1/8int 1/(2x-1)dx + 1/8int 1/(2x+1)dx - 1/2int x/(4x^2+1) dx
To take the integral of each, apply substitution method.
For the first integral, let the substitution be:

u=2x-1
du=2dx
(du)/2=dx

For the second integral, let the substitution be:

v = 2x+1
dv=2dx
(dv)/2=dx

And for the third integral, let the substitution be:

w=4x^2+1
dw=8xdx
(dw)/8=xdx

So the three integrals become:
= 1/8 int 1/u * (du)/2 + 1/8 int 1/v *(dv)/2 -1/2 int 1/w * (dw)/8
=1/16 int 1/u du + 1/16 int 1/v dv - 1/16 int 1/w dx
Then, apply the formula int 1/x dx = ln|x| + C .
=1/16 ln|u| + 1/16 ln|v| - 1/16ln|w|+C
And substitute back u = 2x - 1 , v = 2x + 1 and w = 4x^2+1 .
= 1/16ln|2x-1| + 1/16ln|2x+1| -1/16|ln4x^2+1|+C

Therefore, int x/(16x^4-1)dx=1/16ln|2x-1| + 1/16ln|2x+1| -1/16|ln4x^2+1|+C .